Question

The coordinate of a particle in meters is given by x(t)=16t−3.0t , where the time t is in seconds. The particle is momentarily at rest at t is:

a) 9.3 s
b) 1.3 s
c) 0.75 s
d) 5.3 s

Answer 1

This suggests a potential error either in the initial function describing the particle's position or in the provided answer alternatives.

The accurate solution is not included in the given choices.

Step-by-step explanation:

To find when the particle is momentarily at rest, we need to determine the time at which the velocity
`(\(v(t)\))` is equal to zero. The velocity is the first derivative of the position function \(x(t)\).

Given
`\(x(t) = 16t - 3.0t^2\)`, let's find the velocity
`\(v(t)\)` by taking the derivative of x with respect to t:

`\[v(t) = (dx)/(dt) = (d)/(dt)(16t - 3.0t^2)\]`

`\[v(t) = 16 - 6.0t\]`

Now, set v(t) to zero and solve for t:

`\[16 - 6.0t = 0\]`

`\[6.0t = 16\]`

`\[t = (16)/(6.0) \approx 2.67 \, \text{s}\]`

Therefore, the correct answer is not among the provided options. It seems there might be a mistake in the original function for the particle's position or a typographical error in the answer choices.